Problem 26 - University Physics with Modern Physics

Question

A lens forms an image of an object. The object is 16.0 \(\mathrm{cm}\) from the lens. The image is 12.0 \(\mathrm{cm}\) from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 \(\mathrm{mm}\) tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

Step-by-Step Solution

Answer

(a) Focal length is -48.0 cm; lens is diverging. (b) Image height is 6.375 mm, and it is erect.

Step 1: Understand the Given Information

We are given that an object is 16.0 cm from the lens (\(d_o = 16.0 \, \text{cm}\)). The image is formed 12.0 cm on the same side as the object (\(d_i = -12.0 \, \text{cm}\)). The negative sign indicates that the image is virtual. The object height is 8.50 mm (\(h_o = 8.50 \, \text{mm}\)). The task is to find the focal length, determine the type of lens, calculate the image height, and its nature, as well as to draw a ray diagram.

Step 2: Calculate the Focal Length

Use the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Substitute the given values:\[ \frac{1}{f} = \frac{1}{16.0} + \frac{1}{(-12.0)} \]Calculate:\[ \frac{1}{f} = \frac{1}{16.0} - \frac{1}{12.0} = \frac{3}{48} - \frac{4}{48} = \frac{-1}{48} \]Thus:\[ f = -48.0 \, \text{cm} \]Since the focal length is negative, the lens is diverging.

Step 3: Calculate the Image Height

Use the magnification formula:\[ m = \frac{h_i}{h_o} = \frac{-d_i}{d_o} \]Substitute the given values:\[ m = \frac{-(-12.0)}{16.0} = \frac{12}{16} = 0.75 \]So:\[ h_i = m \times h_o = 0.75 \times 8.50 = 6.375 \, \text{mm} \]The image is erect because the magnification is positive.

Step 4: Describe the Ray Diagram

For a diverging lens, draw parallel rays from the top of the object to the lens. These rays diverge after passing through the lens. Draw another line from the top of the object through the center of the lens, which continues in a straight path. The apparent intersection point of these rays on the object's side forms the image. Since the image is upright and smaller, this confirms it as a virtual image on the same side as the object.

Key Concepts

Focal Length Calculation

The focal length of a lens is a key characteristic that describes how the lens converges or diverges light. In our exercise, we use the lens formula to calculate the focal length:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
This formula relates the focal length \(f\), the object distance \(d_o\), and the image distance \(d_i\). It's essential to pay attention to the sign conventions. Here, the image distance is given as negative because the image forms on the same side as the object, indicating it's virtual.
Substituting the given values:

• \(d_o = 16.0 \, \text{cm}\)
• \(d_i = -12.0 \, \text{cm}\)

We find:
\[ \frac{1}{f} = \frac{1}{16.0} - \frac{1}{12.0} = \frac{-1}{48} \]
The focal length \(f\) is calculated as \(-48.0 \, \text{cm}\). The negative result confirms that the lens at hand is a diverging lens. Diverging lenses spread light rays apart, resulting in a negative focal length.

Magnification and Image Height

Magnification provides insight into how the size of an image relates to the original object. To find the image height, we use the magnification formula:
\[ m = \frac{h_i}{h_o} = \frac{-d_i}{d_o} \]
Where:

• \(h_i\) is the image height
• \(h_o = 8.50 \, \text{mm}\) is the object height
• \(d_i\) and \(d_o\) are the image and object distances, respectively

Substituting the values, we get:
\[ m = \frac{-(-12.0)}{16.0} = \frac{12}{16} = 0.75 \]
The positive magnification of 0.75 indicates that the image is upright and smaller than the object. The image height \(h_i\) is calculated as:
\[ h_i = m \times h_o = 0.75 \times 8.50 = 6.375 \, \text{mm} \]
This confirms the nature of the image: it is erect and diminished, appearing 6.375 mm tall.

Ray Diagrams

Ray diagrams are a visual representation of how light rays travel through lenses. They help us determine where and how the image is formed. For a diverging lens, like the one in our exercise, the following steps characterize the process:

• Draw a parallel ray from the top of the object to the lens. Once it passes through the lens, this ray diverges.
• Next, draw a ray that travels from the top of the object straight through the center of the lens. This ray does not change direction.

The divergent rays seem to originate from a point on the object's side. This point marks where the image is formed. Since these rays do not actually converge to form a real meet, the image is virtual.
Virtual images produced by diverging lenses are upright and appear on the same side as the object, consistent with our exercise, confirming the image's erect state and reduced size.